# NCERT Solutions for Exercise 6.4 Class 12 Maths Chapter 10 - Application of Derivatives

NCERT solutions for exercise 6.4 Class 12 Maths chapter 6 discuss problems related to the approximation of certain quantities using differentiation. Prior to exercise 6.4 Class 12 Maths, there are three exercises in which the concepts of rate of change of quantities, increasing and decreasing function and tangents and normals are discussed. Till the NCERT solutions for Class 12 Maths chapter 6 exercise 6.4 NCERT Book presents 25 solved examples. These NCERT syllabus solved examples give an insight into the topics covered in the chapter. The Class 12 Maths chapter 6 exercise 6.4 gives a detailed explanation of numerical related to the approximation of quantities. As mentioned, other than Class 12th Maths chapter 6 exercise 6.4 there are 5 exercises including the miscellaneous exercise.

Application of Derivatives Exercise 6.1

Application of Derivatives Exercise 6.2

Application of Derivatives Exercise 6.3

Application of Derivatives Exercise 6.5

Application of Derivatives Miscellaneous Exercise

** ****Application of Derivatives Class 12 Chapter 6**** Exercise 6.4**

**Application of Derivatives Class 12 Chapter 6**

**Exercise 6.4**

** Question:1(i) ** Using differentials, find the approximate value of each of the following up to 3

places of decimal.

** Answer: **

Lets suppose and let x = 25 and

Then,

Now, we can say that is approximate equals to dy

Now,

Hence, is approximately equals to 5.03

** Question:1(ii) ** Using differentials, find the approximate value of each of the following up to 3 places of decimal.

** Answer: **

Lets suppose and let x = 49 and

Then,

Now, we can say that is approximately equal to dy

Now,

Hence, is approximately equal to 7.035

** Question:1(iii) ** Using differentials, find the approximate value of each of the following up to 3 places of decimal.

** Answer: **

Lets suppose and let x = 1 and

Then,

Now, we cam say that is approximately equals to dy

Now,

Hence, is approximately equal to 0.8

** Question:1(iv) ** Using differentials, find the approximate value of each of the following up to 3

places of decimal.

** Answer: **

Lets suppose and let x = 0.008 and

Then,

Now, we cam say that is approximately equals to dy

Now,

Hence, is approximately equal to 0.208

** Question:1(v) ** Using differentials, find the approximate value of each of the following up to 3

places of decimal.

** Answer: **

Lets suppose and let x = 1 and

Then,

Now, we cam say that is approximately equals to dy

Now,

Hence, is approximately equal to 0.999 (because we need to answer up to three decimal place)

** Question:1(vi) ** Using differentials, find the approximate value of each of the following up to 3

places of decimal.

** Answer: **

Let's suppose and let x = 16 and

Then,

Now, we can say that is approximately equal to dy

Now,

Hence, is approximately equal to 1.969

** Question:1(vii) ** Using differentials, find the approximate value of each of the following up to 3

places of decimal.

** Answer: **

Lets suppose and let x = 27 and

Then,

Now, we can say that is approximately equal to dy

Now,

Hence, is approximately equal to 2.963

** Question:1(viii) ** Using differentials, find the approximate value of each of the following up to 3

places of decimal.

** Answer: **

Let's suppose and let x = 256 and

Then,

Now, we can say that is approximately equal to dy

Now,

Hence, is approximately equal to 3.997

** Question:1(ix) ** Using differentials, find the approximate value of each of the following up to 3

places of decimal.

** Answer: **

Let's suppose and let x = 81 and

Then,

Now, we can say that is approximately equal to dy

Now,

Hence, is approximately equal to 3.009

** Question:1(x) ** Using differentials, find the approximate value of each of the following up to 3

places of decimal.

** Answer: **

Let's suppose and let x = 400 and

Then,

Now, we can say that is approximately equal to dy

Now,

Hence, is approximately equal to 20.025

** Question:1(xi) ** Using differentials, find the approximate value of each of the following up to 3

places of decimal.

** Answer: **

Lets suppose and let x = 0.0036 and

Then,

Now, we can say that is approximately equal to dy

Now,

Hence, is approximately equal to 0.060 (because we need to take up to three decimal places)

** Question:1(xii) ** Using differentials, find the approximate value of each of the following up to 3

places of decimal.

** Answer: **

Lets suppose and let x = 27 and

Then,

Now, we cam say that is approximately equals to dy

Now,

Hence, is approximately equal to 0.060 (because we need to take up to three decimal places)

** Question:1(xiii) ** Using differentials, find the approximate value of each of the following up to 3

places of decimal.

** Answer: **

Lets suppose and let x = 81 and 0.5

Then,

Now, we can say that is approximately equal to dy

Now,

Hence, is approximately equal to 3.004

** Question:1(xiv) ** Using differentials, find the approximate value of each of the following up to 3

places of decimal.

** Answer: **

Let's suppose and let x = 4 and

Then,

Now, we can say that is approximately equal to dy

Now,

Hence, is approximately equal to 7.904

** Question:1(xv) ** Using differentials, find the approximate value of each of the following up to 3

places of decimal.

** Answer: **

Lets suppose and let x = 32 and

Then,

Now, we can say that is approximately equal to dy

Now,

Hence, is approximately equal to 2.001

** Question:2 ** Find the approximate value of f (2.01), where

** Answer: **

Let x = 2 and

We know that is approximately equal to dy

Hence, the approximate value of f (2.01), where is 28.21

** Question:3 ** Find the approximate value of f (5.001), where

** Answer: **

Let x = 5 and

We know that is approximately equal to dy

Hence, the approximate value of f (5.001), where

** Question:4 ** Find the approximate change in the volume V of a cube of side x metres caused by increasing the side by 1%.

** Answer: **

Side of cube increased by 1% = 0.01x m

Volume of cube =

we know that is approximately equal to dy

So,

Hence, the approximate change in volume V of a cube of side x metres caused by increasing the side by 1% is

** Question:5 ** Find the approximate change in the surface area of a cube of side x metres

caused by decreasing the side by 1%.

** Answer: **

Side of cube decreased by 1% = -0.01x m

The surface area of cube =

We know that, is approximately equal to dy

Hence, the approximate change in the surface area of a cube of side x metres

caused by decreasing the side by 1%. is

** Question:6 ** If the radius of a sphere is measured as 7 m with an error of 0.02 m, then find the approximate error in calculating its volume.

** Answer: **

Error in radius of sphere = 0.02 m

Volume of sphere =

Error in volume

Hence, the approximate error in its volume is

** Question:7 ** If the radius of a sphere is measured as 9 m with an error of 0.03 m, then find the approximate error in calculating its surface area.

** Answer: **

Error in radius of sphere = 0.03 m

The surface area of sphere =

Error in surface area

Hence, the approximate error in its surface area is

** Question:8 ** If , then the approximate value of f (3.02) is

(A) 47.66 (B) 57.66 (C) 67.66 (D) 77.66

** Answer: **

Let x = 3 and

We know that is approximately equal to dy

Hence, the approximate value of f (3.02) is 77.66

Hence, (D) is the correct answer

** Question:9 ** The approximate change in the volume of a cube of side x metres caused by increasing the side by 3% is

(A) 0.06 (B) 0.6 (C) 0.09 (D) 0.9

** Answer: **

Side of cube increased by 3% = 0.03x m

The volume of cube =

we know that is approximately equal to dy

So,

Hence, the approximate change in volume V of a cube of side x metres caused by increasing the side by 3% is

Hence, (C) is the correct answer

**More About NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.4**

Nine questions and their explanations are given in exercise 6.4 Class 12 Maths solutions. There are two objectives questions in the NCERT solutions for Class 12 Maths chapter 6 exercise 6.4. The first question of Class 12 Maths chapter 6 exercise 6.4 have 15 sub questions. All these questions are detailed in the Class 12th Maths chapter 6 exercise 6.4

**Also Read| **Application of Derivatives Class 12 Notes

**Benefits of NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.4**

Exercise 6.4 Class 12 Maths gives more clarity about the approximation technique using differentiation for certain quantities

One question from Class 12 Maths chapter 6 exercise 6.4 may appear for the Class 12 CBSE Exam.

**Also see-**

NCERT Exemplar Solutions Class 12 Maths Chapter 6

NCERT Solutions for Class 12 Maths Chapter 6

**NCERT Solutions Subject Wise**

NCERT Solutions Class 12 Chemistry

NCERT Solutions for Class 12 Physics

NCERT Solutions for Class 12 Biology

NCERT Solutions for Class 12 Mathematics

**Subject Wise NCERT Exemplar Solutions**

NCERT Exemplar Class 12th Maths

NCERT Exemplar Class 12th Physics

NCERT Exemplar Class 12th Chemistry

NCERT Exemplar Class 12th Biology